A038509 -id:A038509 - cp's OEIS frontend

A007310 Numbers congruent to 1 or 5 mod 6.

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175

Offset: 1

C. Christofferson (Magpie56(AT)aol.com)

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)
Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004
Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007
Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016
A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008
Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008
For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008
A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009
Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010
If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010
A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010
Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011
From Peter Bala, May 02 2018: (Start)
The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1.  Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013
(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013
Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014
Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015
a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015
Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016
This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016
The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018
The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020
Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021
From Flávio V. Fernandes, Aug 01 2021: (Start)
For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)
Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023
If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

			G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...

K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016-2018.
B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers.
L. B. W. Jolley, Summation of Series, Dover, 1961
Cedric A. B. Smith, Prime factors and recurring duodecimals, Math. Gaz. 59 (408) (1975) 106-109.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N).
Eric Weisstein's World of Mathematics, Rough Number.
Eric Weisstein's World of Mathematics, Pi Formulas. [_Jaume Oliver Lafont_, Oct 23 2009]
Index entries for sequences related to smooth numbers [_Michael B. Porter_, Oct 09 2009]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).

Filtered([1..150],n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018

a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016

seq(seq(6*i+j,j=[1,5]),i=0..100); # Robert Israel, Sep 08 2014

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *)
a[n_] := (6 n + (-1)^n - 3)/2; a[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)
Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)
Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)

isA007310(n) = gcd(n,6)==1 \\ Michael B. Porter, Oct 09 2009

A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014

\\ given an element from the sequence, find the next term in the sequence.
nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016

def A007310(n): return (n+(n>>1)<<1)-1 # Chai Wah Wu, Oct 10 2023

[i for i in range(150) if gcd(6,i) == 1] # Zerinvary Lajos, Apr 21 2009

a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012
1 - 1/5 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
From Antti Karttunen, May 20 2017: (Start)
a(A273669(n)) = 5*a(n) = A084967(n).
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)

A133900 a(n) = period of the sequence {b(m), m>=0}, defined by b(m):=binomial(m+n,n) mod n.

Original entry on oeis.org

1, 4, 9, 16, 25, 72, 49, 64, 81, 400, 121, 864, 169, 784, 675, 256, 289, 2592, 361, 1600, 1323, 3872, 529, 3456, 625, 5408, 729, 3136, 841, 324000, 961, 1024, 9801, 18496, 6125, 31104, 1369, 23104, 13689, 32000, 1681, 254016, 1849, 15488, 30375, 33856

Offset: 1

Hieronymus Fischer, Oct 15 2007, Oct 20 2007

nonn

This is the analog of the sequence of Pisano periods (A001175) for binomial factors.
n^2 always divides a(n).
A prime p is a factor of a(n) if and only if it is a factor of n (i.e., a(n) and n have the same prime factors).

			a(3)=9 since binomial(m+3,3) mod 3, m>=0, is periodic with period length 3^2=9 (see A133883).
a(6)=72 since binomial(m+6,6) mod 6, m>=0, is periodic with period length 4*6^2=72 (see A133886).

Hieronymus Fischer, Table of n, a(n) for n = 1..111

Cf. A000040, A001175, A133872-A133880, A133882-A133890, A133910.
Cf. A133620-A133625, A133630, A038509, A133634-A133636.
Cf. A133905.

a(n)=n^2 if n is a prime or a power of a prime.

A133620 Binomial(n+p,n) mod n where p=10.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 2, 6, 2, 6, 1, 2, 1, 10, 5, 7, 1, 12, 1, 15, 18, 12, 1, 12, 21, 14, 4, 12, 1, 28, 1, 29, 1, 18, 6, 5, 1, 20, 14, 10, 1, 14, 1, 34, 15, 24, 1, 3, 8, 16, 18, 27, 1, 34, 23, 16, 1, 30, 1, 16, 1, 32, 17, 57, 40, 56, 1, 1, 47, 60, 1, 54, 1, 38, 36, 58, 12, 66, 1, 63, 10, 42, 1

Offset: 1

Hieronymus Fischer, Sep 30 2007

Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.

Ray Chandler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, order 58060800.

Cf. A000040, A133621-A133625, A133630, A038509, A133634, A133635, A133636.

Cf. A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Table[Mod[Binomial[n + 10, n], n], {n, 90}] (* Harvey P. Dale, Apr 04 2015 *)

a(n) = binomial(n+10, n) % n \\ Michel Marcus, Jul 15 2013

a(n) = binomial(n+p,p) mod n.
a(n) = 1 if n is a prime > p, since binomial(n+p,n)==(1+floor(p/n))(mod n), provided n is a prime.
a(n) = A001287(n+10) mod n. - Michel Marcus, Jul 15 2013; corrected by Michel Marcus, Jan 27 2020
For n > 58060802, a(n) = 2*a(n-29030400) - a(n-58060800). - Ray Chandler, Apr 29 2023

A133872 Period 4: repeat [1, 1, 0, 0].

Original entry on oeis.org

1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1

Offset: 0

Hieronymus Fischer, Oct 10 2007

Partial sums of A056594.
Let i=sqrt(-1) and S(n) = Sum_{k=0..n-1} exp(2*Pi*i*k^2/n) for n>=1 the famous Gauss sum. Then S(n) = (a(n)+a(n+1)*i)*sqrt(n). - Franz Vrabec, Nov 08 2007
a(A042948(n)) = 1; a(A042964(n)) = 0. - Reinhard Zumkeller, Oct 03 2008
a(n) is also the real part of partial sum of powers of the complex unit i. - Enrique Pérez Herrero, Aug 16 2009
Periodic sequences having a period of 2k and composed of k ones followed by k zeros have a closed formula of floor(((n+k) mod 2k)/k). Listed sequences of this form are: k=1..A000035(n+1), k=2..A133872(n), k=3..A088911, k=4..A131078(n), k=5..A112713(n-1). - Gary Detlefs, May 17 2011
0.repeat(0,0,1,1) is 1/5 in base 2, due to 1/5 = (3/16)/(1-1/16). For the general case see 1/A062158(n) in base n >= 2. Here n = 2. - Wolfdieter Lang, Jun 20 2014
a(n) (for n>=1) is the determinant of the n X n Toeplitz matrix M satisfying: M(i,j)=1 if -1<=j-i<=2 and 0 otherwise. - Dmitry Efimov, Jun 23 2015
a(n) (for n>=1) is the difference between numbers of even and odd permutations p of 1,2,...,n such that -1 <= p(i)-i <= 2 for i=1,2,...,n. - Dmitry Efimov, Jan 08 2016
The binomial transform is 1, 2, 3, 4, 6, 12,... (see A038504). - R. J. Mathar, Feb 25 2023

			G.f. = 1 + x + x^4 + x^5 + x^8 + x^9 + x^12 + x^13 + x^16 + x^17 + x^20 + ...

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Clark Kimberling, A Combinatorial Classification of Triangle Centers on the Line at Infinity, J. Int. Seq., Vol. 22 (2019), Article 19.5.4.
Psychedelic Geometry Blogspot, Curious Series-001 [_Enrique Pérez Herrero_, Aug 16 2009]
Index entries for characteristic functions
Index entries for linear recurrences with constant coefficients, signature (1,-1,1).

Cf. A056594, A133620-A133625, A133630, A038509, A133634-A133636, A021913, A000217, A133882, A133880, A133890, A133900, A133910, A000035, A088911, A131078, A112713.

[ (1 + (-1)^Floor(n/2))/2 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

&cat[[1,1,0,0]^^25]; // Vincenzo Librandi, Jan 09 2016

A133872:=n->(1+(-1)^((2n-1+(-1)^n)/4))/2;
seq(A133872(n), n=0..100); # Wesley Ivan Hurt, Dec 06 2013

Table[(1 + (-1)^((2 n - 1 + (-1)^n)/4))/2, {n, 0, 100}] (* Wesley Ivan Hurt, Dec 06 2013 *)
PadRight[{},120,{1,1,0,0}] (* Harvey P. Dale, Jan 26 2014 *)

a(n)=n%4<2 \\ Jaume Oliver Lafont, Mar 17 2009

Vec((1+x)/(1-x^4) + O(x^100)) \\ Altug Alkan, Jan 08 2015

def A133872(n): return int(not n&2) # Chai Wah Wu, Jan 31 2023

maxn <- 63 # by choice
a <- c(1,0,0)
for(n in 4:maxn) a[n] <- a[n-1] - a[n-2] + a[n-3]
(a <- a(1,a))
# Yosu Yurramendi, Oct 25 2020

a(n) = (1 + floor(n/2)) mod 2.
a(n) = A004526(A000035(n+2)).
a(n) = 1 + floor(n/2) - 2*floor((n+2)/4).
a(n) = (((n+2) mod 4) - (n mod 2))/2.
a(n) = ((n + 2 - (n mod 2))/2) mod 2.
a(n) = ((2*n + 3 + (-1)^n)/4) mod 2.
a(n) = (1 + (-1)^((2*n - 1 + (-1)^n)/4))/2.
a(n) = binomial(n+2, n) mod 2 = binomial(n+2, 2) mod 2.
a(n) = A000217(n+1) mod 2.
G.f.: (1+x)/(1-x^4) = 1/((1-x)(1+x^2)).
a(n) = 1/2 + (1/2)*cos(Pi*n/2) + (1/2)*sin(Pi*n/2). a(n) = A021913(n+2). - R. J. Mathar, Nov 15 2007
From Jaume Oliver Lafont, Dec 05 2008: (Start)
a(n) = 1/2 + sin((2n+1)Pi/4)/sqrt(2).
a(n) = 1/2 + cos((2n-1)Pi/4)/sqrt(2). (End)
a(n) = Re(Sum_{k=0..n} i^k), where i=sqrt(-1) and Re is the real part of a complex number. a(n) = (1/2)*((Sum_{k=0..n} i^k) + Sum_{k=0..n} i^-k) = Re((1/2)*(1 + i)*(1 - i^(n+1))). - Enrique Pérez Herrero, Aug 16 2009
a(n) = (1 + i^(n*(n-1)))/2, where i=sqrt(-1). - Bruno Berselli, May 18 2011
a(n) = (Sum_{k=1..n} k^j) mod 2, for any j. - Gary Detlefs, Dec 28 2011
a(n) = a(n-1) - a(n-2) + a(n-3) for n>2. - Jean-Christophe Hervé, May 01 2013
a(n) = 1 - floor(n/2) + 2*floor(n/4) = 1 - A004526(n) + A122461(n). - Wesley Ivan Hurt, Dec 06 2013
a(n) = (1 + (-1)^floor(n/2))/2. - Wesley Ivan Hurt, Apr 17 2014
a(n) = A054925(n+2) - A011848(n+2). - Wesley Ivan Hurt, Jun 09 2014
Euler transform of length 4 sequence [1, -1, 0, 1]. - Michael Somos, Sep 26 2014
a(n) = a(1-n) for all n in Z. - Michael Somos, Sep 26 2014
From Ilya Gutkovskiy, Jul 09 2016: (Start)
Inverse binomial transform of A038504(n+1).
E.g.f.: (exp(x) + sin(x) + cos(x))/2. (End)
a(n) = (1 + (-1)^(n*(n-1)/2))/2. - Guenther Schrack, Apr 04 2019

Definition rewritten by N. J. A. Sloane, Apr 30 2009

A133890 Binomial(n+10,n) mod 10.

Original entry on oeis.org

1, 1, 6, 6, 1, 3, 8, 8, 8, 8, 6, 6, 6, 6, 6, 0, 5, 5, 0, 0, 5, 5, 0, 0, 0, 6, 6, 6, 6, 6, 8, 8, 3, 3, 8, 6, 1, 1, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 5, 5, 6, 6, 1, 1, 6, 8, 8, 8, 8, 8, 6, 6, 6, 6, 1, 5, 0, 0, 5, 5, 0, 0, 0, 0, 0, 6, 6, 6, 6, 6, 3, 3, 8, 8, 3, 1, 6, 6, 6, 6, 0, 0, 0, 0, 0, 0, 5, 5, 0, 0, 1, 1, 6, 6, 6

Offset: 0

Hieronymus Fischer, Oct 10 2007

Periodic with length A133900(10)=4*10^2=400.

Ray Chandler, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, order 231.

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636, A133880, A133900, A133910.

Table[Mod[Binomial[10+n,n],10],{n,0,120}] (* Harvey P. Dale, Sep 19 2018 *)

a(n)=binomial(n+10,10)%10 \\ Charles R Greathouse IV, Nov 22 2011

a(n) = binomial(n+10,10) mod 10.

A133630 Nonprime numbers k such that binomial(k+p,k) mod k = 1, where p=10.

Original entry on oeis.org

4, 33, 57, 68, 85, 87, 111, 121, 141, 143, 164, 169, 185, 187, 209, 219, 221, 235, 247, 249, 253, 260, 289, 292, 299, 303, 319, 323, 327, 335, 341, 356, 361, 377, 381, 388, 391, 403, 407, 411, 435, 437, 451, 452, 473, 481, 484, 485, 489, 493, 516, 517, 519

Offset: 1

Hieronymus Fischer, Sep 30 2007

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636, A133880, A133890, A133900, A133910.

is(n)=binomial(n+10,10)%n==1 && !isprime(n) \\ Charles R Greathouse IV, Nov 04 2015

A133880 n modulo p repeated p times (where p=10).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1

Offset: 0

Hieronymus Fischer, Oct 10 2007

Periodic with length p^2=100.

a(n) = A179051(n) for n < 90. - Reinhard Zumkeller, Jun 27 2010

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1).

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.

Cf. A133890, A133900, A133910.

Flatten[Table[PadRight[{},10,Mod[n,10]],{n,11}]] (* Harvey P. Dale, May 10 2012 *)

a(n)=(n\10+1)%10 \\ Charles R Greathouse IV, Nov 15 2022

The following formulas are given for a general parameter p (p=10 for this sequence).
a(n)=(1+floor(n/p)) mod p.
a(n)=1+floor(n/p)-p*floor((n+p)/p^2).
a(n)=(((n+p) mod p^2)-(n mod p))/p.
a(n)=((n+p-(n mod p))/p) mod p.
G.f. g(x)=((p-1)x^(p^2)-px^(p(p-1))+1)/((1-x)(1-x^p)(1-x^(p^2))).
G.f. g(x)=(1-x^p)*sum{0<=k<(p-1), (k+1)*x^(k*p)}/((1-x)(1-x^(p^2))).

A133625 Binomial(n+p, n) mod n where p=5.

Original entry on oeis.org

0, 1, 2, 2, 2, 0, 1, 7, 4, 3, 1, 8, 1, 8, 9, 13, 1, 7, 1, 10, 8, 12, 1, 3, 6, 1, 10, 8, 1, 2, 1, 25, 12, 1, 8, 22, 1, 20, 14, 39, 1, 15, 1, 12, 25, 24, 1, 5, 1, 11, 18, 14, 1, 46, 12, 43, 20, 1, 1, 48, 1, 32, 22, 49, 14, 23, 1, 18, 24, 50, 1, 7, 1, 1, 41, 20, 1, 66, 1, 77, 28, 1, 1, 50, 18, 44

Offset: 1

Hieronymus Fischer, Sep 30 2007

Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.

Ray Chandler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, order 240.

Cf. A000040, A133620-A133625, A133630, A038509, A133634-A133636.

Cf. A133875, A133885, A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.

Table[Mod[Binomial[n+5,n],n],{n,90}] (* Harvey P. Dale, Oct 02 2015 *)

a(n)=binomial(n+5,5) mod n.
a(n)=1 if n is a prime > 5, since binomial(n+5,n)==(1+floor(5/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^5 + 15*n^4 + 85*n^3 + 105*n^2 + 34*n + 120)/120 mod n.
For n > 6:
if n mod 120 == 0, then a(n) = 17*n/60 + 1.
if n mod 120 is in {1, 2, 7, 11, 13, 17, 19, 23, 26, 29, 31, 34, 37, 41, 43, 47, 49, 53, 58, 59, 61, 67, 71, 73, 74, 77, 79, 82, 83, 89, 91, 97, 98, 101, 103, 106, 107, 109, 113, 119}, then a(n) = 1.
if n mod 120 is in {3, 9, 18, 21, 27, 33, 39, 42, 51, 57, 63, 66, 69, 81, 87, 93, 99, 111, 114, 117}, then a(n) = n/3 + 1.
if n mod 120 is in {4, 28, 44, 52, 68, 76, 92, 116}, then a(n) = n/4 + 1.
if n mod 120 is in {5, 10, 25, 35, 50, 55, 65, 85, 95, 115}, then a(n) = n/5 + 1.
if n mod 120 is in {6, 54, 78, 102}, then a(n) = 5*n/6 + 1.
if n mod 120 is in {8, 16, 32, 56, 64, 88, 104, 112}, then a(n) = 3*n/4 + 1.
if n mod 120 is in {12, 36, 84, 108}, then a(n) = 7*n/12 + 1.
if n mod 120 is in {14, 22, 38, 46, 62, 86, 94, 118}, then a(n) = n/2 + 1.
if n mod 120 is in {15, 45, 75, 90, 105}, then a(n) = 8*n/15 + 1.
if n mod 120 is in {20, 100}, then a(n) = 9*n/20 + 1.
if n mod 120 is in {24, 48, 72, 96}, then a(n) = n/12 + 1.
if n mod 120 == 30, then a(n) = n/30 + 1.
if n mod 120 is in {40, 80}, then a(n) = 19*n/20 + 1.
if n mod 120 == 60, then a(n) = 47*n/60 + 1.
if n mod 120 is in {70, 110}, then a(n) = 7*n/10 + 1.
(End)
For n > 246, a(n) = 2*a(n-120) - a(n-240). - Ray Chandler, Apr 23 2023

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A133633 Duplicate of A038509.

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A133900 a(n) = period of the sequence {b(m), m>=0}, defined by b(m):=binomial(m+n,n) mod n.

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A133620 Binomial(n+p,n) mod n where p=10.

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A133872 Period 4: repeat [1, 1, 0, 0].

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A133890 Binomial(n+10,n) mod 10.

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